I've been struggling with this, I need to find the final velocity of a car after it travels a certain distance when I know it's initial velocity and acceleration are known. Any help?
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$\begingroup$Here's a way forward.
Motion under constant acceleration in one dimension boils down to these equations:
$$x(t) = x(0) + v(0)t + at^2/2$$ $$v(t) = v(0) + at$$
Here, $x(t)$ is the displacement as a function of time $t$, $v(t)$ is the velocity as a function of time, and $a$ is the constant acceleration. Additionally, $x(0), v(0)$ can be interpreted as your initial displacement and initial velocity.
(To get these two equations, one would start with $dv/dt = a$ and integrate twice with respect to $t$.)
Solve the second equation for $t$, and substitute in the first to eliminate $t$.
Can you take it from here?
$\endgroup$ 2 $\begingroup$Perhaps try looking at the problem this way:
$\int_a^b{A(t)}=V(b) - V(a)$; where $a$=initial time, $b$=final time
This shows that a definite integral of acceleration is the difference in final velocity from initial velocity. If you are given the velocity and acceleration as functions of $t$, this approach is probably your best bet. Hopefully this helps!
$\endgroup$ $\begingroup$Have you considered the basic Kinematic equations?
Let:
- $V_f$ denote the (unknown) final velocity of the car.
- $V_0$ denote the (known) initial velocity of the car.
- $a$ denote the (known) acceleration of the car.
- $d$ denote the certain distance.
You can now proceed multiple ways, but i'll just recommend using the Kinematic equation $$V_f^2=V_0^2+2\times a\times d$$ Simply plug in the knowns and it should be very straightforward.
Note: This is the simple case, where the acceleration is constant and we consider only one dimension, which is what I assume you are looking for?
An Example: Say you are given that the car has initial velocity of $V_0=5$ $m/s$, and a constant acceleration of $a=2$ $m/s^2$. You want the final velocity $V_f$ of the car after it travels $d=4$ meters. Using the above Kinematic equation , you have: $$V_f^2=5^2+2\times2 \times 4$$ so $V_f^2=25+16=41$. Take the square root of both sides and you get $V_f \approx 6.403$ m/s.
$\endgroup$ $\begingroup$The general solution for the distance traveled by an object while speed varies between $v_1$ and $v_2$ with variable acceleration is
$$ x = \int \limits_{v_1}^{v_2} \frac{v}{a} \,{\rm d}v $$
In your case the acceleration is constant and the distance is given
$$ x = \frac{v_2^2-v_1^2}{2 a} $$
which is solved for $v_2$.
Note also that $t =\int \frac{1}{a} \,{\rm d}v$, $t =\int \frac{1}{v} \,{\rm d}x$ and $\frac{1}{2} (v_2^2-v_1^2) = \int a \,{\rm d}x$
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